Year: 2007
Paper: 3
Question Number: 2
Course: LFM Stats And Pure
Section: Generalised Binomial Theorem
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item Show that $1.3.5.7. \;\ldots \;.(2n-1)=\dfrac {(2n)!}{2^n n!}\;$
and that, for $\vert x \vert
< \frac14$,
\[
\frac{1}{\sqrt{1-4x\;}\;}
=1+\sum_{n=1}^\infty \frac {(2n)!}{(n!)^2} \, x^n \,.
\]
\item By differentiating the above result, deduce that
\[
\sum _{n=1}^\infty \frac{(2n)!}{n!\,(n-1)!}
\left(\frac6{25}\right)^{\!\!n} =
60
\,.
\]
\item Show that
\[
\sum _{n=1}^\infty \frac{2^{n+1}(2n)!}{3^{2n}(n+1)!\,n!}
=
1
\,.
\]
\end{questionparts}\begin{questionparts}
\item Notice that $1 \cdot 3 \cdot 5 \cdot 7 \cdot (2n-1) = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots \cdots 2n}{2 \cdot 4 \cdot 6 \cdots 2n} = \frac{(2n)!}{2^n \cdot n!}$ as required.
When $|4x| < 1$ or $|x|<\frac14$ we can apply the generalised binomial theorem to see that:
\begin{align*}
\frac{1}{\sqrt{1-4x}} &= (1-4x)^{-\frac12} \\
&= 1+\sum_{n=1}^\infty \frac{-\frac12 \cdot \left ( -\frac32\right)\cdots \left ( -\frac{2n-1}2\right)}{n!} (-4x)^n \\
&= 1+\sum_{n=1}^{\infty} (-1)^n\frac{(2n)!}{(n!)^2 2^{2n}} (-4)^n x^n \\
&= 1+\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2 } x^n \\
\end{align*}
\item Differentiating we obtain \begin{align*}
&& 2(1-4x)^{-\frac32} &= \sum_{n=1}^\infty \frac{(2n)!}{n!(n-1)!} x^{n-1} \\
\Rightarrow &&\sum_{n=1}^\infty \frac{(2n)!}{n!(n-1)!} \left (\frac{6}{25} \right)^{n} &= \frac{6}{25} \cdot 2\left(1- 4 \frac{6}{25}\right)^{-\frac32} \\
&&&= \frac{12}{25} \left (\frac{1}{25} \right)^{-\frac32} \\
&&&= \frac{12}{25} \cdot 125 = 60
\end{align*}
\item By integrating, we obtain
\begin{align*}
&& \int_{t=0}^{t=x} \frac{1}{\sqrt{1-4t}} \d t &= \int_{t=0}^{t=x} \left (1+\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2 } t^n \right) \d t \\
\Rightarrow && \left [ -\frac12 \sqrt{1-4t}\right]_0^x &= x + \sum_{n=1}^{\infty} \frac{(2n)!}{n!(n+1)! } x^{n+1} \\
\Rightarrow && \frac12 - \frac12\sqrt{1-4x} - x &= \sum_{n=1}^{\infty} \frac{(2n)!}{n!(n+1)! } x^{n+1} \\ \\
\Rightarrow && 9\sum_{n=1}^{\infty} \frac{(2n)!}{n!(n+1)! } \left ( \frac{2}{9}\right)^{n+1} &=9 \cdot\left( \frac12 - \frac12 \sqrt{1-4\cdot \frac{2}{9}} - \frac29\right )\\
&&&= 9 \cdot \left (\frac12 - \frac{1}{6} - \frac{2}{9} \right) \\
&&&= 1
\end{align*}
\end{questionparts}
This question was popular though not well answered. Solutions to part (i) were frequently unconvincing, though to part (ii) were quite good if they avoided elementary errors in working. Part (iii) was less well attempted with some not spotting to use integration, some stumbling over "+ c" and some not spotting the value of x to substitute.